Question

A bracket connected to the flange of a column through a group of rivets to transfer a load of 120 kN at an eccentricity of 150 mm as shown in the figure. The thickness of the bracket plate is 8 mm. Design the diameter of the rivets if the permissible stress in single shear is 100 MPa and in bearing 180 MPa.

Answer 1

$F_{d_i}=\text{direct shear force in the}{i}^{th}\text{rivet}$

Since all the rivets are of same dia

$\therefore F_{d_i}=\frac{p}{n}=\frac{120}{10}=12kN$

$P_{T_i}=\text{Torsional shear force in the}{i}^{th}\text{rivet.}$

$F_{T_i}=\frac{\left(Pe\right)r_i}{\sum r_i^2}$

$\text{Rivet (1) is critical as}{F}_{T_i}\text{is highest for that rivet}$

$r_1=\sqrt{{45}^2+{120}^2}=128.16mm$

$\sum r_i^2=\sum x^2+\sum y^2$
= $10\times {45}^2+4\times {120}^2+4\times {60}^2$
= 92250 $m{m}^2$

$\therefore F_{T_1}=\frac{120\times 150\times {10}^3\times 128.16}{92250}=25kN$

Resultant force on rivet (1) is
$F_{r_1}=\sqrt{F_{d_1}^2+F_{T_1}^2+2F_{d1}\times F_{T_1}\times \cos \Theta }$
$F_{r_1}$ = $\sqrt{{12}^2+{25}^2+\frac{2\times 12\times 25\times 45}{128.16}}$

$F_{r_1}$ = 31299.75 N

Now, we don't know the rivet value of Rivets. So, we will find dia of rivet by considering both bearing and shearing and maximum dia will be adopted.

Shearing

$100\times \frac{\pi }{4}\times d^2=31299.75$

$\therefore$ d = 19.96 mm


Bearing
$180\times d\times 8=31299.75$
$\therefore$ d = 21.73 mm
Let us provide rivets of dia 22 mm.